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Bạn đã thử cố gắng giải nó chưa..
Ta có :
$\cos (\pi +x)=\cos ((x-\pi)+2\pi)=\cos (x-\pi)=\cos (\pi -x )=-\cos x$
$\sin(\frac{3\pi+x}{2} )=\sin (\frac{x}{2}-\frac{\pi}{2}+2\pi)=\sin (\frac{x}{2}-\frac{\pi}{2})=-\sin (\frac{\pi}{2}-\frac{x}{2})=-\cos \frac{x}{2}$
Như vậy PT $\Leftrightarrow 1+\cos x+\cos \frac{x}{2}=0 (*)$
Bây giờ chú ý rằng $ \cos x=2\cos^2 \frac{x}{2}-1$.
PT $(*)\Leftrightarrow 2\cos^2 \frac{x}{2}+\cos \frac{x}{2}=0\Leftrightarrow \cos \frac{x}{2}\left (2\cos \frac{x}{2}+1\right )=0$
$\Leftrightarrow \left[ {\begin{matrix} \cos \frac{x}{2}=0\\ \cos \frac{x}{2}=-\frac{1}{2} \end{matrix}} \right.$$\Leftrightarrow \left[ {\begin{matrix} x=\pi + 2k\pi\\ x=\pm \frac{4\pi}{3}+4k\pi \end{matrix}} \right. (k \in \mathbb{Z}).$
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